Answer
(a) $F_3 = 2.9~\hat{j}~N$
(b) $F_4 = -1.6~\hat{i}~N$
Work Step by Step
(a) $F_1+F_{2y}+ F_3 = 0$
$F_3 = -F_1-F_{2y}$
$F_3 = [-(-5.0~\hat{j})-(6.0)~sin(20^{\circ})~\hat{j}]~N$
$F_3 = 2.9~\hat{j}~N$
(b) $F_{2x}+F_4= 4.0~\hat{i}~N$
$F_4= 4.0~\hat{i}~N-F_{2x}$
$F_4= 4.0~\hat{i}~N-(6.0)~cos(20^{\circ})~\hat{i}~N$
$F_4 = -1.6~\hat{i}~N$
